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Ksenya-84 [330]
3 years ago
10

An ideal spring is fixed at one end. A variable force F pulls on the spring. When the magnitude of F reaches a value of 43.8 N,

the spring is stretched by 15.5 cm from its equilibrium length. Calculate the additional work required by F to stretch the spring by an additional 10.4 cm from that position. (in J)
Physics
1 answer:
maw [93]3 years ago
5 0

A force of 43.8 N is required to stretch the spring a distance of 15.5 cm = 0.155 m, so the spring constant <em>k</em> is

43.8 N = <em>k</em> (0.155 m)   ==>   <em>k</em> = (43.8 N) / (0.155 m) ≈ 283 N/m

The total work done on the spring to stretch it to 15.5 cm from equilibrium is

1/2 (283 N/m) (0.155 m)² ≈ 3.39 J

The total work needed to stretch the spring to 15.5 cm + 10.4 cm = 25.9 cm = 0.259 m from equilibrium would be

1/2 (283 N/m) (0.259 m)² ≈ 9.48 J

Then the additional work needed to stretch the spring 10.4 cm further is the difference, about 6.08 J.

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